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2016大连网络赛 Sparse Graph

2024-08-12 22:24:53   219人阅读

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)


Problem Description
In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G.

Now you are given an undirected graph G of N nodes and M bidirectional edges of unit length. Consider the complement of G, i.e., H. For a given vertex S on H, you are required to compute the shortest distances from S to all N−1 other vertices.
 

 

Input
There are multiple test cases. The first line of input is an integer T(1≤T<35) denoting the number of test cases. For each test case, the first line contains two integers N(2≤N≤200000) and M(0≤M≤20000). The following M lines each contains two distinct integers u,v(1≤u,v≤N) denoting an edge. And S (1≤S≤N) is given on the last line.
 

 

Output
For each of T test cases, print a single line consisting of N−1 space separated integers, denoting shortest distances of the remaining N−1 vertices from S (if a vertex cannot be reached from S, output ``-1" (without quotes) instead) in ascending order of vertex number.
 

 

Sample Input
12 01
 

 

Sample Output
1
 
分析：对于能够到达的点依次放入队列，暴力未放入的点，可行继续放入队列即可；
代码：
#include <iostream>#include <cstdio>#include <cstdlib>#include <cmath>#include <algorithm>#include <climits>#include <cstring>#include <string>#include <set>#include <map>#include <queue>#include <stack>#include <vector>#include <list>#define rep(i,m,n) for(i=m;i<=n;i++)#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)#define mod 1000000007#define inf 0x3f3f3f3f#define vi vector<int>#define pb push_back#define mp make_pair#define fi first#define se second#define ll long long#define pi acos(-1.0)#define pii pair<int,int>#define Lson L, mid, rt<<1#define Rson mid+1, R, rt<<1|1const int maxn=2e5+10;const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};using namespace std;ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}int n,m,k,t,dp[maxn];set<int>a,b[maxn],c;queue<int>p;int main(){    int i,j;    scanf("%d",&t);    while(t--)    {        bool flag=false;        scanf("%d%d",&n,&m);        memset(dp,-1,sizeof dp);        rep(i,1,n)b[i].clear(),a.insert(i);        while(m--)        {            int u,v;            scanf("%d%d",&u,&v);            b[u].insert(v),b[v].insert(u);        }        scanf("%d",&m);        p.push(m);dp[m]=0;a.erase(m);        while(!p.empty())        {            c.clear();            int u=p.front();p.pop();            for(int x:a)if(b[x].find(u)==b[x].end())dp[x]=dp[u]+1,c.insert(x),p.push(x);            for(int x:c)a.erase(x);        }        rep(i,1,n)if(i!=m)        {            if(flag)printf(" %d",dp[i]);            else printf("%d",dp[i]),flag=true;        }        printf("\n");    }    //system("pause");    return 0;}

2016大连网络赛 Sparse Graph


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